Slope change 1 in 95 to 1 in 1420 with jump

Given: wide rectangular channel, $q = 3.75\ \text{m}^2/\text{s}$, $n = 0.013$, slopes $1/95 = 0.01053$ and $1/1420 = 0.000704$; $R \approx y$.

Normal and critical depths

$$y_n = \left(\frac{nq}{\sqrt{S_0}}\right)^{3/5},\qquad y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{3.75^2}{9.81}\right)^{1/3} = 1.128\ \text{m}$$ $$y_{n1} = \left(\frac{0.013 \times 3.75}{\sqrt{0.01053}}\right)^{0.6} = 0.640\ \text{m}\quad (< y_c)$$ $$y_{n2} = \left(\frac{0.013 \times 3.75}{\sqrt{0.000704}}\right)^{0.6} = 1.440\ \text{m}\quad (> y_c)$$

The upstream slope is steep with supercritical uniform flow, $V_1 = 5.86$ m/s and $Fr_1 = 2.34$; the downstream slope is mild with subcritical uniform flow. The transition must be a hydraulic jump.

Location and height

Sequent depth of $y_{n1}$:

$$y_2 = \frac{0.640}{2}\left(\sqrt{1 + 8(2.34)^2} - 1\right) = 1.821\ \text{m}$$

This is more than $y_{n2} = 1.440$ m, so the downstream depth is insufficient to force the jump at the junction. The jump forms on the mild slope, downstream of the junction, after an M3 curve in which the depth rises from 0.640 m. The jump begins at the depth $y_j$ whose sequent depth equals 1.440 m:

$$y_j:\ \frac{y_j}{2}\left(\sqrt{1 + 8\frac{q^2}{g\,y_j^3}} - 1\right) = 1.440 \;\Rightarrow\; y_j = 0.864\ \text{m}$$

Height of the jump $= y_{n2} - y_j = 1.440 - 0.864 = 0.577$ m.

Direct step for the M3 length (3 steps from 0.640 to 0.864 m, $S_0 = 0.000704$, $S_f = n^2q^2/y^{10/3}$): $\Delta x = 33.2 + 32.1 + 30.1 \approx 95$ m. So the jump is about 95 m downstream of the junction.

Sketch

   steep 1/95            |       mild 1/1420
 -- yn1 = 0.640 ---------+--.                      
                         |   '-.  M3     J    _____ yn2=1.440
                         |      '-..____/'|---'
                         |   0.640 -> 0.864 | 1.440

Answer: $y_{n1} = 0.640$ m, $y_{n2} = 1.440$ m; a jump from 0.864 m to 1.440 m (height 0.58 m) forms about 95 m downstream of the junction, preceded by an M3 curve.